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Brrunno [24]
1 year ago
14

Solve y=x+5 and 2x-3y=4

Mathematics
2 answers:
Lyrx [107]1 year ago
8 0

\huge \tt \color{pink}{A}\color{blue}{n}\color{red}{s}\color{green}{w}\color{grey}{e}\color{purple}{r }

  • To solve such problem there are two method by elimination method or substitution method
  • I have solved with substitution method juz' refer to attachment for understanding the ans !

{ \boxed{ \: ✟\underline{  \boxed{ \sf{ \: x =  - 19 ✓}}}✟ \:}} \\ { \boxed{ \: ✟\underline{  \boxed{ \sf{ \: y =  - 14✓ }}}✟ \:}} \\\\

Hope it helps !

stepan [7]1 year ago
4 0

Answer:

x= -19

y= -14

Step-by-step explanation:

Substitution.

It tells you what y is already. You substitute that into the second equation and then solve. Then you use that answer and substitute it back into the first equation's x, and then you'll get your full answer.

Hope this helps <3

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Negative one fourths times negative six elevenths
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You multiply fractions simply by multiplying numerators and denominators with each other:

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Perform the indicated row operations, then write the new matrix.
Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]

Operations:

2R_1 + R_2 ->R_2

-3R_1 +R_3 ->R_3

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

Step-by-step explanation:

The first operation:

2R_1 + R_2 ->R_2

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by 2

2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}0&5&7|1\end{array}\right]

The second operation:

-3R_1 +R_3 ->R_3

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by -3

-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]

Hence, the new matrix is:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

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